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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    One kg of water, at \[20{}^\text{o}C,\]is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of \[20\Omega \]. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to :
    [Specific heat of water \[=4200\text{ }J/kg\text{ }{}^\text{o}C\]),
    Latent heat of water = 2260 kJ/kg]
    [JEE Main 12-4-2019 Afternoon]

    A) 3 minutes                    

    B) 22 minutes

    C) 10 minutes                  

    D) 16 minutes

    Correct Answer: B

    Solution :

    \[Q=P\times t\] \[Q=mc\Delta T+mL\] \[P=\frac{V_{rms}^{2}}{R}\] \[4200\times 80+2260\times {{10}^{3}}=\frac{{{(200)}^{2}}}{20}\times t\] \[t=1298\text{ }sec\] \[t\simeq 22\min \]


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